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(a) How many ordered pairs (x,y) of integers are there such that \sqrt{x^2 + y^2} = 5? Does the question have a geometric interpretation? 

(b) How many ordered triples (x,y,z) of integers are there such that \sqrt{x^2 + y^2 + z^2} = 7? Does the question have a geometric interpretation?

 

I know these questions have been asked before but the explinations are a little hard to follow and don't have a clear answer.  Any help would be great!  

 

EDIT: Okay I got the number of pairs but I don't know how to turn it into a geometric interpretation.  I got a: 12 and b: 54 btw

 Jun 26, 2019
edited by Guest  Jun 26, 2019
 #1
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(a)

 

\(\sqrt{x^2+y^2}\ =\ 5\)

 

The solutions to this equation are all points with a distance of  5  from the origin.

 

So this is the equation of a circle with a radius of 5 centered at the origin.

 

 

I used a little help from here: https://math.stackexchange.com/questions/518856/integral-points-on-a-circle

 

There is a Pythaogrean triple with a hypotenuse of 5

A triangle with side lengths 3, 4 and 5 is a Pythagorean triple. So...

 

From the origin, we can go over 3 units and either up or down 4 units to reach an integer solution.

From the origin, we can go over 4 units and either up or down 3 units to reach an integer solution.

 

From the origin, we can go over 0 units and either up or down 5 units to reach an integer solution.

From the origin, we can go over 5 units and either up or down 0 units to reach an integer solution.

 

Here's a graph showing all integer solutions:   https://www.desmos.com/calculator/txwzj5nmt4

 

There are  12  integer solutions.

 Jun 26, 2019
 #2
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b? Do I need to make a 3d graph?

 Jun 27, 2019
 #3
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(b) How many ordered triples (x,y,z) of integers are there such that \sqrt{x^2 + y^2 + z^2} = 7? Does the question have a geometric interpretation?

 

I  get the following  triples

 

6, 3, 2    

   

But each of these can take on pos/ neg   values for each integer   so there are 2^3  = 8 possibilities  and each of these can be arranged in 3! ways  = 6 ways....so  8 * 6  = 48  possibilities

 

And

 

7 0  0       the "7" can be pos/negative   = 2 possibilities  and can be in any 3 positions  = 2 * 3  = 6 possibilities

 

So.....(If I haven't missed any)  we have 48 + 6   =  54  points [ just as you found ]

 

This is a sphere centered at  (0, 0, 0)   with a radius of 7

 

Here's a (not so good) image : 

 

 

cool cool cool

 Jun 27, 2019

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